Suppose {tex}\\text{α,β and γ}{/tex} are the zeros of the said polynomial p(x)Then, we have\xa0{tex}\\alpha{/tex}\xa0= 2,\xa0{tex}\\beta{/tex}\xa0= -3 and\xa0{tex}\\gamma{/tex}\xa0= 4Now,{tex}\\text{α+β +γ=2-3+4=3}{/tex}……(1){tex}\\text{αβ +βγ+γα=2(-3)+(-3)(4)+(4)(2)=-6-12+8=-10……(.2)}{/tex}{tex}\\text{αβγ=2(-3)(4)=-24 …….(3)}{/tex}Now, a cubic polynomial whose zeros are\xa0{tex}\\alpha, \\space \\beta \\space and \\space \\gamma{/tex}\xa0is given byp(x) = x3\xa0-\xa0{tex}\\text{(α+β+γ)x}^2{/tex}\xa0+\xa0{tex}\\text{(αβ+βγ+γα)x}{/tex}\xa0-\xa0{tex}\\alpha\\beta\\gamma{/tex}Now putting the values from (1),(2) and (3) we getp(x) = x3\xa0-(3)x2\xa0+ (-10)x – (-24)= x3\xa0-3×2 -10x +24